7/28/2005 SECOND HOURLY PRACTICE Maths21a, O. Knill, Summer 2005Name:• Start by printing your name in the above box.• Try to answer each question on the same page as the question is asked. If needed, usethe back or the next empty page fo r work. If you need additional paper, write your nameon it.• Do not detach pages from this exam packet or unstaple the packet.• Please write neatly. Answers which are illegible for the grader can not be given credit.• No notes, books, calculators, computers, or other electronic aids can be allowed.• You have 90 minutes time to complete your work.1 202 103 104 105 106 107 108 109 10Total: 100Problem 1) (10 points)T FThe speed of the curve ~r(t) = (cos(t), sin(t), 3t) is (−sin(t), cos(t), 3).Solution:This is the velocity. The speed is the length of this vector and would be√10.T FEvery smooth function of three variables f(x, y, z) satisfies the partial dif-ferential equation fxyz+ fyzx= 2fzxy.Solution:By Clairot’s theorem.T FIf fx(x, y) = fy(x, y) for all x, y, then f(x, y) is a constant.Solution:The transport equation fx= fyis a PDE with solutions like for example f(x, y) = x + y.Any function which stays invariant by replacing x with y is a solution: like f (x, y) =sin(xy) + x5y5.T F(1, 1) is a local maximum of the function f(x, y) = x2y −x + cos(y).Solution:(1, 1) is not even a critical point.T FIf f is a smoo th function of two variables, then the number of critical pointsof f inside the unit disc is finite.Solution:Take f(x, y) = x2for example. Every point on the y a xes {x = 0} is a critical point.T FThe value of the function f(x, y) = sin(−x + 2y) at (0.001, −0.002) can bylinear approximation be estimated as −0.003.Solution:The correct approximation would be f(0, 0) + 0.001(−1) − 0.002(2) = −0.005.T FIf (1, 1) is a critical point for the f unction f(x, y) then (1, 1) is also a criticalpoint for the function g(x, y) = f(x2, y2).Solution:If ∇f(1, 1) = (fx(1, 1), fy(1, 1)) = (0, 0) then also ∇g(1, 1) = (fx(1, 1)2x, fy(1, 1)2y) =(0, 0). Note that the statement would not be true, if we would replace (1, 1) say with(2, 2) (as in the practice exam).T FIf the velocity vector ~r′(t) of the planar curve ~r(t) is orthogonal t o thevector ~r(t) for a ll times t, then the curve is a circle.Solution:d/dt(~r(t) ·~r(t)) = ~r(t) ·~r′(t).T FThe curvature of a circle of radius 10 is 1/10.Solution:Yes, the curvature of a circle of radius r is 1/10.T FThe arc length of a curve is given by the formulaRba~r′(t) dt.Solution:Take absolute values.T FThe gradient of f(x, y) is normal to the level curves of f .Solution:This is a basic and important fact.T FIf (x0, y0) is a maximum of f(x, y) under the constraint g(x, y) = g(x0, y0),then (x0, y0) is a maximum of g(x, y) under the constraint f(x, y) =f(x0, y0).Solution:Assume you have a situation f, g, where this is true and where the constraint is g = 0,produce a new situation f, h = −g, where the first statement is still true but where t heextrema of h under the constraint of f is a minimum.T FIf ~u is a unit vector tangent at (x, y, z) to the level surface of f(x, y, z) t henthe directional derivative satisfies Duf(x, y, z) = 0.Solution:The directional derivative measures the rate of change of f in the direction of u. On alevel surface, in the direction of the surface, the f unction does not change (because f isconstant by definition on the surface).T FIf ~r(t) = (x(t), y(t)) and x(t), y(t) are polynomials, then the tangent line isdefined at all points.Solution:Take the example r(t) = (t2, t3). At t = 0, we have a cusp and the gradient is zero. Wedo not have a tangent line there.T FThe function u(x, t) = x2/2 + t satisfies the heat equation ut= uxx.Solution:Just differentiate.T FThe vector ~ru(u, v) is tangent to the surface parameterized by ~r(u, v) =(x(u, v), y(u, v), z(u, v)).Solution:The vector ~ruis tangent to a grid curve and so tangent to the surface.T FClairots theorem implies fxyx= fyxySolution:We have fxy= fyxbut in the formula, we have to the left two x derivatives and to theright two y derivatives.T FThe second derivative test allows to check whether an extremum found withthe Lagrange multiplyer method is a maximum.Solution:No, the second derivative test applies for function f(x, y) without constraint.T FIf (0, 0) is a critical point of f(x, y) and the discriminant D is zero butfxx(0, 0) > 0 then (0, 0) can not be a local maximum.Solution:If fxx(0, 0) > 0 then on the x-axes the function g(x) = f(x, 0) has a local minimum. Thismeans that there are points close to (0, 0) where the value of f is larger.T FLet (x0, y0) be a saddle point of f (x, y). For any unit vector ~u, there a r epoints arbitrarily close to (x0, y0) for which ∇f is parallel to ~u.Solution:Just look at the level curves near a saddle point. The gradient vectors are orthogonalto the level curves which a re hyperbola. You see that they point in any direction except4 directions. To see this better, take a pen and draw a circle around the saddle pointbetween two of your knuckles on your fist. At each point of the circle, now draw thedirection of steapest increase (this is the gradient direction).Problem 2) (10 points)Match the contour maps with the corresponding functions f (x, y) of two variables. Note thatone of the contour maps is not represented by a formula. No justifications are needed.I II IIIIV V VIEnter I,II,I II,IV,V or VI here Function f(x, y)f(x, y) = sin(x)f(x, y) = x2+ 2y2f(x, y) = |x| + |y|f(x, y) = xe−x2−y2f(x, y) = x2/(x2+ y2)Solution:Enter I,II,I II,IV,V or VI here Function f(x, y)IIIf(x, y) = sin(x)IIf(x, y) = x2+ 2y2Vf(x, y) = |x| + |y|If(x, y) = xe−x2−y2VIf(x, y) = x2/(x2+ y2)Match the Parametrizations with the curves. No justifications are needed.I II IIIIV V VIEnter I,II,..., until VI here Equation~r(t) = (cos(t), sin(t))~r(t) = (cos(3t), sin(5t))~r(t) = (t cos(3t), t sin(3t))~r(t) = (t2, t3− t5)~r(t) = (cos(t)2, sin(t)2)~r(t) = (t2, t3)Solution:V I II III IV VIProblem 3) (10 points)a) Use the technique of linear approximation to estimate f(log(2) + 0.001, 0.006) for f(x, y) =e2x−y. (Here, log means the natural logar ithm).b) Find the equation ax + by = d for the tangent line which goes through the point (log(2), 0).Solution:a) L(x, y) = f (x0, y0) + fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0)f(x0, y0) = e2 log 2= 4fx(x0, y0) = 8fy(x0, y0) = −4L(x, y) = 4 + 0.0 01 · 8 −4 · 0.006 =3.984 .b) We have a = 8 and b = −4 and get d = 8 log(2 ) so that the line has
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